List all files in a directory in order of timestamp

have files from directory displayed in an array of columns,using the list icon. however i would like them to be in order of aquisition or timestamp, rather then all 1, 10ns 20s etc.
is there anyway to control the indexing with time stamping order?

LabVIEW allows you to sort just about anything. All you have to do is create a cluster containing everything you want for a "record" and make an array of those clusters. In your case, you want to cluster the file name with its time stamp (how about last modification date?). The sort function (we call them functions or VIs, not icons) will order your array based on the order in which you clustered your elements.
I whipped up a tiny example. Give it a directory path and run it. The output is a list of the filed in that folder sorted according to their time stamp.
If you have some clear specifications for your whole application, consider hiring an Alliance Member (like us). You will end up with a most professional example of LabVIEW code t
o learn from and tweak as you like.
Alliance Members are here to help. We do this stuff every day.
Daniel L. Press
PrimeTest Corp.
www.primetest.com
Attachments:
List_Files_by_Time.vi ‏21 KB

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List all files in a directory on a server that hosts the applet

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So i tried it like this
File fi = new URL(getCodeBase() + "/all/").getFile();But when I try to list the files, I get a SecurityException.
Can anyone help me out?
Peace!
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http://search.java.sun.com/search/java/index.jsp?col=javaforums&qp=&qt=%2Blist+%2Bfile+%2Bserver

List all files in a directory

How can I list all the files in a directory which has folders and files as well? I just wanna files..
File dataDir = new File("c:/data/);
File[] listing = dataDir.listFiles();
but it list the immediate no of folders in the data directory. Is there a way to get all the files in the subdirectories in data folder and excluding the folders?
thanks!

Iterate over the returned array, ignoring those whose isDirectory() returns true.
Alternatively, you could call listFiles(fileNameFilter) instead, with an implementation of FileNameFilter which rejects File objects whose isDirectory() returns true.

List all files in a directory, not including the sub directories if any

Hi,
I have been looking around php.net for a bit and can not work
out how i list the files that are in a directory i.e.
www.site.com/directory/
I would like the names of each file to be placed in an array,
but not to have the sub directories in this list should there be
any.
I was given some code a while ago that done this but it
listed the sub directories and i would like them not in this list.
I do not have this code anymore and do not know where i got
it, so i can not get it amended to what i need.
please can someone tell me what line of code i should be
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thank you in advance for your help.

(_seb_) wrote:
> not very clever wrote:
>> Hi,
>>
>> I have been looking around php.net for a bit and can
not work out how
>> i list the files that are in a directory i.e.
>>
>> www.site.com/directory/
>>
>> I would like the names of each file to be placed in
an array, but not
>> to have the sub directories in this list should
there be any.
>>
>> I was given some code a while ago that done this but
it listed the
>> sub directories and i would like them not in this
list.
>>
>> I do not have this code anymore and do not know
where i got it, so i
>> can not get it amended to what i need.
>>
>> please can someone tell me what line of code i
should be using.
>>
>> thank you in advance for your help.
>>
>>
>>
>
> The follwoing PHP code will do just that. Just replace
"pathToFolder"
> with the path to your folder.
> I made the list of files also link to each file. Just
remove the link
> echo if you don't them to be links.
>
> <?php
> // FUNCTION TO LIST FILES:
> function listFiles($path){
> if($handle = opendir($path)){
> while(false !== ($file = readdir($handle))){
> if (is_file($path.'/'.$file) &&
!preg_match('/^\./',$file)){
> $files_array[]=$file;
> }
> }
> }
> return($files_array);
> }
> $path = 'pathToFolder';
>
> // CALL THE FUNCTION:
> $files_array = listFiles($path);
> foreach($files_array as $file){
> echo '<a
href="'.$path.'/'.$file.'">'.$file,'</a>';
> }
>
> ?>
>
I spotted one error:
foreach($files_array as $file){
echo '<a
href="'.$path.'/'.$file.'">'.$file,'</a>';
should be:
foreach($files_array as $file){
echo '<a
href="'.$path.'/'.$file.'">'.$file.'</a>';
(a dot after $file, not a coma)
seb ( [email protected])
http://webtrans1.com | high-end web
design
Downloads: Slide Show, Directory Browser, Mailing List

How to list all files and directories in another directory

I need to be able to list all the directories and files in a directory. I need to write a servlet that allows me to create an html page that has a list of files in that directory and also list all the directories. That list of files will be put into an applet tag as a parameter for an applet that I have already written. I am assuming that reading directories/files recursively on a web server will be the same as reading directories/files on a local system, but I don't know how to do that either.

Hi,
Here is a method to rotate through a directory and put all the files into a Vector (files).
 * Iterates throught the files in the root file / directory that is passed
 * as a parameter. The files are loaded into a <code>Vector</code> for
 * processing later.
 * @param file the root directory or the file
 * that you wish to have inspected.
 public void loadFiles(File file) {
 if (file.isDirectory()) {
 File[] entry= file.listFiles();
 for (int i= 0; i < entry.length; i++) {
 if (entry.isFile()) {
 //Add the file to the list
 files.add(entry[i]);
 } else {
 if (entry[i].isDirectory()) {
 //Iterate over the entries again
 loadFiles(entry[i]);
 } else {
 if (file.isFile()) {
 //Add the file
 files.add(file);
See ya
Michael

How to get all files in one directory

Hi there,
is there any way to get all the files in one directory?
e.g. a method
File[] getAllFiles(String directory){
Thank you

Just out of interest, which part of the File API was confusing?
http://java.sun.com/j2se/1.5.0/docs/api/index.html
If you look down the left-hand side, you'll see a frame listing "All File". If you select "File" from that list, it opens up the API in the right-hand frame. Quite near the top you'll find "listFiles".
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How can I create a array with all files from a directory

How can I create a array of files or varchar with all files from a directory?

I thought the example could be improved upon. I've posted a solution on my blog that doesn't require writing the directory list to a table. It simply returns it as a nested table of files as a SQL datatype. You can find it here:
http://maclochlainn.wordpress.com/2008/06/05/how-you-can-read-an-external-directory-list-from-sql/

List of files in a directory of app server

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How to get the file size (in bytes) for all files in a directory?

How to get the file size (in bytes) for all files in a directory?
The following code does not work. isFile() does NOT recognize files as files but only as directories. Why?
Furthermore the size is not retrieved correctly.
How do I have to code it otherwise? Is there a way of not converting f-to-string-to-File again but iterate over all file objects instead?
Thank you
Peter
java.io.File f = new java.io.File("D:/todo/");
files = f.list();
for (int i = 0; i < files.length; i++) {
System.out.println("fn=" + files);
if (new File(files[i]).isFile())
 System.out.println("file[" + i + "]=" + files[i] + " size=" + (new File(files[i])).length() ); }

pstein wrote:
...The following code does not work. Work?! It does not even compile! Please consider posting code in the form of an SSCCE in future.
Here is an SSCCE.
import java.io.File;
class ListFiles {
 public static void main(String[] args) {
 java.io.File f = new java.io.File("/media/disk");
 // provides only the file names, not the path/name!
 //String[] files = f.list();
 File[] files = f.listFiles();
 for (int i = 0; i < files.length; i++) {
 System.out.println("fn=" + files);
if (files[i].isFile()) {
System.out.println(
"file[" +
i +
"]=" +
files[i] +
" size=" +
(files[i]).length() );
}Edit 1:
Also, in future, when posting code, code snippets, HTML/XML or input/output, please use the code tags to retain the indentation and formatting. To do that, select the code and click the CODE button seen on the Plain Text tab of the message posting form. It took me longer to clean up that code and turn it into an SSCCE, than it took to +solve the problem.+
Edited by: AndrewThompson64 on Jul 21, 2009 8:47 AM

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Message was edited by:
Oli_t

Creating list of files from a directory

Hi,
poked around on the internet, and wanted to be able to create an ArrayList of all files in a directory. This code compiles, but doesn't work:
import java.io.*;
import java.util.*;
public class TestDir {
 public static void main(String[] args){
 File folder = new File("testdir");
 System.out.println("does the folder exist? " + folder.exists());
 System.out.println("is folder a directory? " + folder.isDirectory());
 System.out.println("path: " + folder.getAbsolutePath());
 // create list of files
 try{
 ArrayList<File> files = TestDir.getFileList(folder);
 Iterator i = files.iterator();
 while (i.hasNext()){
 File temp = (File) i.next();
 System.out.println("file: " + temp.getName());
 } catch (IOException e){
 System.err.println("problem creating file list:");
 e.printStackTrace();
 } // end main
 public static ArrayList<File> getFileList(File dir) throws FileNotFoundException{
 ArrayList<File> result = new ArrayList<File>();
 File[] files = dir.listFiles();
 List<File> tempfiles = Arrays.asList(files);
 for(File file : files){
 result.add(file);
 return result;
} // end classthis compiles fine, but doesn't give me the list of files.
Any idea what I'm doing wrong?
bp

console output:
C:\jason\java\domPractice>java TestDir
does the folder exist? true
is folder a directory? true
path: C:\jason\java\domPractice\testdir

Getting the list of files in a directory by their last modified date

Dear friends,
I want to get the list of files in a directory sorted by their last modified date.
By default the file.list() or listFiles() return files sorted by their name in ascending order.
Please give me your suggestions.
Thanks in advance,
James.

Thanks friend,
I myself got the answer
here is my code:
public File[] getSortedFileList(File dir){
File[] originalList = dir.listFiles();
int numberOfFiles = originalList.length;
File[] sortedList = new File[numberOfFiles];
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for(int i = 0; i < numberOfFiles; i++){
lastModified[i] = originalList.lastModified();
Arrays.sort(lastModified);
for(int i = 0; i < numberOfFiles; i++){
for(int k = 0; k < numberOfFiles; k++){
if(originalList[k].lastModified() == lastModified[i])
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[Bash] Massively replace text in all files of a directory

Hi everybody,
I wrote this small recursive function in order to massively replace some strings contained in all files of a directory (and all subdirectories). Any suggestions?
replaceText() {
# set the temporary location
local tFile="/tmp/out.tmp.$$"
# call variables
local script="$2"
local opts="${@:3}"
browse() {
for iFile in "$1"/*; do
if [ -d "$iFile" ];then
# enter subdirectory...
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elif [ -f $iFile -a -r $iFile ]; then
echo "$iFile"
sed $opts "s/$(echo $script)/g" "$iFile" > $tFile && mv $tFile "$iFile"
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browse $1
Syntax:
replaceText [path] [script] [sed options (optional)]
For example (it will replace "hello" with "hi" in all files):
replaceText /home/user/mydir hello/hi
Note: It is case-sensitive.
Bye,
grufo
Last edited by grufo (2012-11-10 15:05:43)

falconindy wrote:
Yes, find is recursive and extremely good at its job.
http://mywiki.wooledge.org/UsingFind
Well
falconindy wrote:Your lack of quoting is dangerous, as is your code injection in sed. I'm not sure why you're echoing a var inside a command substitution inside a sed expression, but it's going to be subject to word splitting, all forms of expansion, and may very well break the sed expression entirely, leading to bad things. A contrived example, but passing something like 'foo//;d;s/bar/' should effectively delete the contents of every file the function touches.
So, if you consider it dangerous, you can adopt the whole "sed syntax" and confirm before continue...:
replaceText() {
# set the temporary location
local tFile="/tmp/out.tmp.$$"
# call variables
local sedArgs="${@:2}"
browse() {
for iFile in "$1"/*; do
if [ -d "$iFile" ];then
# enter subdirectory...
browse "$iFile"
elif [ -f $iFile -a -r $iFile ]; then
echo "$iFile"
sed $sedArgs "$iFile" > $tFile && mv $tFile "$iFile"
else
echo "Skip $iFile"
fi
done
while true; do
read -p "Do you want to apply \"sed $sedArgs\" to all files contained in the directory $1? [y/n] " yn
case $yn in
[Yy]* ) browse $1; break;;
* ) exit;;
esac
done
Syntax:
replaceText [parent directory] [sed arguments]
Example:
replaceText /your/path -r 's/OldText/NewText/g'
or, if you want to work directly with the current directory...
replaceText() {
# set the temporary location
local tFile="/tmp/out.tmp.$$"
# call variables
local sedArgs="$@"
browse() {
for iFile in "$1"/*; do
if [ -d "$iFile" ];then
# enter subdirectory...
browse "$iFile"
elif [ -f $iFile -a -r $iFile ]; then
echo "$iFile"
sed $sedArgs "$iFile" > $tFile && mv $tFile "$iFile"
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done
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replaceText [sed arguments]
Example:
replaceText -r 's/OldText/NewText/g'
What about?
falconindy wrote:I'll also point out that declaring a function within a function doesn't provide any amount of scoping -- 'browse' will be declared in the user's namespace after running this function for the first time.
See:
function1() {
function2() {
echo "Ciao"
function2
function2 # error
function1 # works

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List all files in a directory in order of timestamp

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